Question

Se considera functia f:R cu valori in R, f(x) =2 x la a treia - 9x la a treia +12x+1.
a) Aratati ca f'(x) = 6 (x-1) (x-2),x apartine lui R
b) calculati lim (x->+0) 2x la a treia -f(x) supra f'(x)
c) determinati ecuatia tangentei la graficul funciei f in punctul de abscisa x=1,situat pe graficul functiei f

Answer 1

[tex]f(x)=2 x^{3}-9 x^{2}+12x+1 \\ a).f'(x)=6x^{2} -18x+12=6( x^{2} -3x+2). \\ Ec.atas: x^{2} -3x+2=0 =\ \textgreater \ x1=1;x2=2 \\ =\ \textgreater \ f'(x)=6(x-1)(x-2). \\ b). \lim_{x \to \zer0} \frac{2 x^{3} -(2 x^{3}-9 x^{2} +12x+1) }{6 x^{2} -18x+12}= \lim_{x \to \zer0} \frac{2 x^{3}-2 x^{3}+9 x^{2} -12x-1}{6 x^{2} -18x+12}= \\ \lim_{x \to \zer0} \frac{9 x^{2} -12x-1}{6 x^{2} -18x+12} = \lim_{x \to \zer0} \frac{9*0-12*0-1}{6*0-18*0+12}= -\frac{1}{12}. \\ [/tex].$\\c).x=1; \\ f(1)=2-9+12+2=6 \\ f'(1)=6(1-3+2)=0. \\ Notam: d: y-f(1)=f'(1)(x-1) =\ \textgreater \ d: y-6=0(x-1) \\ =\ \textgreater \ d: y=6$


Sper ca nu am gresit la calcule.